Oil Deposits HDU - 1241 搜索

The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numerous square plots. It then analyzes each plot separately, using sensing equipment to determine whether or not the plot contains oil. A plot containing oil is called a pocket. If two pockets are adjacent, then they are part of the same oil deposit. Oil deposits can be quite large and may contain numerous pockets. Your job is to determine how many different oil deposits are contained in a grid.

Input

The input file contains one or more grids. Each grid begins with a line containing m and n, the number of rows and columns in the grid, separated by a single space. If m = 0 it signals the end of the input; otherwise 1 <= m <= 100 and 1 <= n <= 100. Following this are m lines of n characters each (not counting the end-of-line characters). Each character corresponds to one plot, and is either `*', representing the absence of oil, or `@', representing an oil pocket.

Output

For each grid, output the number of distinct oil deposits. Two different pockets are part of the same oil deposit if they are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than 100 pockets.

Sample Input

1 1
*
3 5
*@*@*
**@**
*@*@*
1 8
@@****@*
5 5 
****@
*@@*@

题目大意

     给你n和m，输入n行m列的字符（字符只能为'@'和'*'），如果一个'@'的相邻单位也是'@'的话，他们算一块油田，最后求出一共有几块油田。

思路

    输入完字符之后，for循环查找，当找到'@'时，广搜（bfs）它的四周，当发现'@'时，将其变为'*'。

代码
 

#include <iostream>

using namespace std;
int n,m;
char s[105][105];
int dz[8][2]={{-1,0},{0,-1},{1,0},{0,1},{-1,-1},{-1,1},{1,-1},{1,1}};
void bfs(int x,int y)
{
    if(s[x][y]!='@')
        return ;
    else
    {
    s[x][y]='*';
    for(int i=0;i<8;i++)
    {
        int x1 =x+dz[i][0];
        int y1 =y+dz[i][1];
        if(x1>=0 &&x1<n &&y1>=0 &&y1<m &&s[x1][y1]=='@')
        {
            bfs(x1,y1);
        }
    }
    }
}
int main()
{
    while(cin>>n>>m)
    {
        if(n==0 &&m==0)
            break;
        int cnt=0;
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
            cin>>s[i][j];
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(s[i][j]=='@')
                {
                    bfs(i,j);
                    cnt++;
                }
            }
        }
        cout<<cnt<<endl;
    }
    return 0;
}